Answer
$f(x)=\sum_{n=0}^{\infty} c_{n}x^{n}=\frac{(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})}{1-x^{4}}$
$R=1$, with interval of convergence of $(-1,1)$
Work Step by Step
$\sum_{n=0}^{\infty} c_{n}x^{n}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+....+$
Separate this series into 4 chunks.
$\sum_{n=0}^{\infty} c_{n}x^{n}=(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})+x^{4}(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})+....$
we get
$f(x)=\sum_{n=0}^{\infty} c_{n}x^{n}=\frac{(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})}{1-x^{4}}$
$R=1$, with interval of convergence of $(-1,1)$
